Q13/042) Solve ∣√(x−1)−2∣+∣√(x−1)−3∣=1

We know (√(x−1)−2)-(√(x−1)−3) = 1
so let (√(x−1)−2 = t

|t | + |t -1 | = 1

t cannot be < 0 then t-1 is less than -1 and  |t | + |t -1 | > 1

t cannot be > 1 then t-1 is less than -1 and  |t | + |t -1 | > 1

so we need to check 0 < = t < = 1

if t = 1 then |t | + |t -1 | = 1

if t = 0 then |t | + |t -1 | = 1

if 0 < t < 1 then t-1 < 0 so |t | + |t -1 | = t  + 1- t = 1

so 0 < = t < = 1

or   0 <= √(x−1)−2 <= 1

or 2 < = √(x−1) or <= 3

 5 < = x  <= 10